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\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 1. Complex Numbers \\ 
Section 1. The Algebra of Complex Numbers}
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle

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\begin{frame}{Contents }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}


\begin{enumerate}

\item  {\color{red}The Algebra of Complex Numbers}

\begin{enumerate}
\item[1.1.] {\color{red}Arithmetic Operations}
\item[1.2.] {\color{red}Square Roots}
\item[1.3.] {\color{red}Justification}
\item[1.4.] {\color{red}Conjugation, Absolute Value}
\item[1.5.] {\color{red}Inequalities}
\end{enumerate}

\item  The Geometric Representation of Complex Numbers

\begin{enumerate}
\item[2.1.] Geometric Addition and Multiplication
\item[2.2.] The Binomial Equation
\item[2.3.] Analytic Geometry
\item[2.4.] The Spherical Representation
\end{enumerate}

\end{enumerate}

\end{frame}


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%\begin{frame}{ 第一章：复数 }
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%\begin{tabular}{|p{1cm}|p{2cm}|p{9cm}|}  \hline 
%章节 &标题 & 重点和难点 \\ \hline 
%1.1 & 复数的 \newline 代数运算 & 1.计算复数的代数运算。%（编程）
%\newline 2.复数开根号。
%\newline 3.复数开任意次根号。
%\newline 4.使用几何作图实现复数的加法和乘法运算。 
%\newline 5.证明复数的三角不等式与柯西不等式。%（难点）
%\newline 6.使用二阶矩阵的加法和乘法来实现复数的加法与乘法。
%\\ \hline 
%1.2 & 复数的 \newline 几何表示 & 7.用复数方程表示平面几何图形。%（编程）
%\newline 8.使用球极投影表示扩展的复平面。%（难点）
%\\ \hline 
%1.A & 编程 & Python 的 cmath 库： \newline 
%\url{https://docs.python.org/3/library/cmath.html}
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%\end{tabular}
%\end{center}
%}
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\begin{frame}{ 作业1A: 1-6 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 复数的代数基本计算。
\item[2.] 复数开根号的计算。
\item[3.] 证明复数域与实数矩阵代数的一个子集同构。
\item[4.] 证明复数形式的拉格朗日恒等式。
\item[5.] 证明三角不等式中的等号成立的充分必要条件。
\item[6.] 证明复数形式的柯西不等式。

\end{enumerate}

\end{frame}

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\begin{frame}{1. The Algebra of Complex Numbers }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
It is fundamental that real and complex numbers obey the same basic laws of arithmetic. 

\item[2.] 
We begin our study of complex function theory by stressing and implementing this analogy.

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.]  
From elementary algebra the reader is acquainted with the {\color{blue}imaginary unit} $i$ with the property $i^2 = -1$. 

\item[2.] 
If the imaginary unit is combined with two real numbers $\alpha, \beta$ by the processes of addition and multiplication, we obtain a complex number $\alpha + i\beta$.

\item[3.] 
$\alpha$ and $\beta$ are the {\color{blue}real and imaginary part} of the complex number. 

\item[4.] 
If $\alpha = 0$, the number is said to be purely imaginary; if $\beta = 0$, it is of course real. 

\item[5.] 
Zero is the only number which is at once real and purely imaginary. 


\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[6.] 
Two complex numbers are equal if and only if they have the same real part and the same imaginary part. 

\item[7.] 
Addition and multiplication do not lead out from the system of complex numbers. 

\item[8.] 
Assuming that the ordinary rules of arithmetic apply to complex numbers we find indeed
\begin{equation}
(\alpha + i \beta) + (\gamma + i\delta) = (\alpha + \gamma) + i(\beta + \delta)
\label{eq-1} 
\end{equation}
and
\begin{equation}
(\alpha + i \beta) (\gamma + i\delta) = (\alpha\gamma - \beta\delta) + i(\alpha\delta + \beta\gamma). 
\label{eq-2}
\end{equation}

\item[9.] 
In the second identity we have made use of the relation $i^2 = -1$. 

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[10.] 
{\color{red}It is less obvious that division is also possible.}

\item[11.] 
We wish to show that $(\alpha + i\beta)/(\gamma + i\delta)$ is a complex number,  provided that $\gamma + i\delta \neq 0$. 

\item[12.] 
If the quotient is denoted by $x + iy$, we must have 
\begin{equation*}
\alpha + i \beta = (\gamma + i\delta) (x + iy).
%\label{eq-}
\end{equation*}

\item[13.] 
By (\ref{eq-2}) this condition can be written
\begin{equation*}
\alpha + i \beta = (\gamma x - \delta y) + i(\delta x + \gamma y), 
%\label{eq-}
\end{equation*}
and we obtain the two equations
\begin{equation*}
\begin{aligned}
\alpha &= \gamma x - \delta y \\  
\beta &= \delta x + \gamma y.  
\end{aligned}
%\label{eq-}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[14.] 
This system of simultaneous linear equations has the unique solution
\begin{equation*}
\begin{aligned}
x &= \frac{\alpha\gamma + \beta\delta}{\gamma^2 + \delta^2} \\  
y &= \frac{\beta\gamma - \alpha\delta}{\gamma^2 + \delta^2}, 
\end{aligned}
%\label{eq-}
\end{equation*}
for we know that $\gamma^2+\delta^2$ is not zero.

\item[15.] 
We have thus the result
\begin{equation}
\frac{\alpha + i \beta}{\gamma + i\delta} 
= \frac{\alpha\gamma + \beta\delta}{\gamma^2 + \delta^2} 
+ \frac{\beta\gamma - \alpha\delta}{\gamma^2 + \delta^2}.
\label{eq-3}
\end{equation}

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[16.] 
Once the existence of the quotient has been proved, its value can be found in a simpler way. 

\item[17.] 
If numerator and denominator are multiplied with $\gamma-i\delta$, we find at once
\begin{equation*}
\frac{\alpha + i \beta}{\gamma + i\delta} 
=\frac{(\alpha + i \beta)(\gamma - i\delta)}{(\gamma + i\delta)(\gamma - i\delta)} 
= \frac{(\alpha\gamma + \beta\delta) + i(\beta\gamma - \alpha\delta)}{\gamma^2 + \delta^2}.
%\label{eq-}
\end{equation*}

\item[18.] 
As a special case the reciprocal of a complex number $\neq 0$ is given by 
\begin{equation*}
\frac{1}{\alpha + i \beta} = \frac{\alpha - i \beta}{\alpha^2 + \beta^2}. 
%\label{eq-}
\end{equation*}

\item[19.] 
We note that $i^n$ has only four possible values: $1, i, -1, -i$. 

\item[20.] 
They correspond to values of $n$ which divided by $4$ leave the remainders $0, 1,
2, 3$.
 

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. Exercise - 1 \hfill 作业1A-1 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the values of
$$
(1+2i)^3, \hspace{0.5cm}
\frac{5}{-3+4i}, \hspace{0.5cm}
\left(\frac{2+i}{3-2i}\right)^2, \hspace{0.5cm}
(1+i)^n+(1-i)^n.
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $z = x + iy$ ($x$ and $y$ real), find the real and imaginary parts of
$$
z^4,  \hspace{0.5cm}
\frac{1}{z},  \hspace{0.5cm}
\frac{z-1}{z+1},  \hspace{0.5cm}
\frac{1}{z^2}. 
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.1. Arithmetic Operations. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that
$$
\left( \frac{-1\pm i\sqrt{3}}{2} \right)^3=1,
\,\,\,\mathrm{and}\,\,\,
\left( \frac{\pm 1\pm i\sqrt{3}}{2} \right)^6=1
$$
for all combinations of signs.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Square Roots. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
{\color{red}We shall now show that the square root of a complex number can be found explicitly. }

\item[2.] 
If the given number is $\alpha + i\beta$, we are looking for a number $x + iy$ such that 
\begin{equation*}
(x + iy)^2 = \alpha + i\beta.
%\label{eq-}
\end{equation*}

\item[3.] 
This is equivalent to the system of equations
\begin{equation}
\begin{aligned}
x^2- y^2 &= \alpha \\ 
2xy &= \beta.
\end{aligned}
\label{eq-4}
\end{equation}

\item[4.] 
From these equations we obtain
\begin{equation*}
(x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2 = \alpha^2 + \beta^2.
%\label{eq-}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{1.2. Square Roots. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
Hence we must have
\begin{equation*}
x^2 + y^2 = \sqrt{\alpha^2 + \beta^2},
%\label{eq-}
\end{equation*}
where the square root is positive or zero. 

\item[6.] 
Together with the first equation (\ref{eq-4}) we find
\begin{equation}
\begin{aligned}
x^2 &= \frac{1}{2}(\alpha + \sqrt{\alpha^2 + \beta^2}) \\ 
y^2 &= \frac{1}{2}(-\alpha + \sqrt{\alpha^2 + \beta^2}). 
\end{aligned}
\label{eq-5}
\end{equation}

\item[7.] 
Observe that these quantities are positive or zero regardless of the sign of $\alpha$.


\end{enumerate}

\end{frame}

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\begin{frame}{1.2. Square Roots. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[8.] 
The equations (\ref{eq-5}) yield, in general, two opposite values for $x$ and two for $y$.

\item[9.] 
But these values cannot be combined arbitrarily, for the second equation (\ref{eq-4}) is not a consequence of (\ref{eq-5}).

\item[10.] 
We must therefore be careful to select $x$ and $y$ so that their product has the sign of $\beta$. 

\item[11.] 
This leads to the general solution
\begin{equation}
\sqrt{\alpha + i\beta} =\pm\left(
\sqrt{\frac{\alpha+\sqrt{\alpha^2 + \beta^2}}{2} }
+ i \frac{\beta}{|\beta|}\sqrt{\frac{-\alpha+\sqrt{\alpha^2 + \beta^2}}{2} }
\right)
\label{eq-6}
\end{equation}
provided that $\beta\neq 0$. 


\end{enumerate}

\end{frame}

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\begin{frame}{1.2. Square Roots. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[12.] 
For $\beta = 0$ the values are $\pm \sqrt{\alpha}$; if $\alpha \ge 0$, $\pm i \sqrt{-\alpha}$ if $\alpha < 0$. 

\item[13.] 
It is understood that all square roots of positive numbers are taken with the positive sign. 

\item[14.] 
We have found that the square root of any complex number exists and has two opposite values. 

\item[15.] 
They coincide only if $\alpha + i\beta = 0$.

\item[16.] 
They are real if $\beta = 0, \alpha \ge 0$ and purely imaginary if $\beta = 0, \alpha \le 0$. 


\end{enumerate}

\end{frame}

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\begin{frame}{1.2. Square Roots. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[17.] 
In other words, except for zero, only positive numbers have real square roots and only negative numbers have purely imaginary square roots. 

\item[18.] 
Since both square roots are in general complex, it is not possible to distinguish between the positive and negative square root of a complex number. 

\item[19.] 
We could of course distinguish between the upper and lower sign in (\ref{eq-6}), but this distinction is artificial and should be avoided. 

\item[20.] 
{\color{blue}The correct way is to treat both square roots in a symmetric manner.}


\end{enumerate}

\end{frame}

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\begin{frame}{1.2. Square Roots. Exercise - 1 \hfill 作业1A-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Compute
$$
\sqrt{i}, \hspace{0.5cm}
\sqrt{-i}, \hspace{0.5cm}
\sqrt{1+i}, \hspace{0.5cm}
\sqrt{\frac{1-i\sqrt{3}}{2}}.
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Square Roots. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the four values of $\sqrt[4]{-1}$. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Square Roots. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Compute $\sqrt[4]{i}$ and $\sqrt[4]{-i}$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Square Roots. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Solve the quadratic equation
$$
z^2 + (\alpha + i\beta)z + \gamma + i\delta = 0.
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
So far our approach to complex numbers has been completely uncritical. 

\item[2.] 
{\color{red}We have not questioned the existence of a number system in which the equation $x^2 + 1 = 0$ has a solution while all the rules of arithmetic remain in force. }

\item[3.] 
We begin by recalling the characteristic properties of the real-number system which we denote by $\mathbb{R}$.

\item[4.] 
In the first place, $\mathbb{R}$ is a {\color{blue}field}. 

\item[5.] 
This means that addition and multiplication are defined, satisfying the associative, commutative, and distributive laws. 


\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[6.] 
The numbers 0 and 1 are neutral elements under addition and multiplication, respectively: $\alpha + 0 = \alpha$, $\alpha \cdot 1 = \alpha$ for all $\alpha$. 

\item[7.] 
Moreover, the equation of subtraction $\beta + x = \alpha$ has always a solution, and the equation of division $\beta x = \alpha$ has a solution whenever $\beta\neq 0$. 

\item[8.] 
One shows by elementary reasoning that the neutral elements and the results of subtraction and division are unique. 

\item[9.] 
Also, every field is an {\color{blue}integral domain}: $\alpha\beta = 0$ if and only if $\alpha = 0$ or $\beta = 0$.

\item[10.] 
These properties are common to all fields. 

\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[11.] 
In addition, the field $\mathbb{R}$ has an {\color{blue}order} relation $\alpha < \beta$ (or $\beta > \alpha$). 

\item[12.] 
It is most easily defined in terms of the set $\mathbb{R}^+$ of positive real numbers: $\alpha < \beta$ if and only if $\beta - \alpha \in \mathbb{R}^+$.

\item[13.] 
{\color{red}The set $\mathbb{R}^+$ is characterized by the following properties: }
\begin{enumerate}
\item[(1)] {\color{red}0 is not a positive number; }
\item[(2)] {\color{red}if $\alpha \neq 0$ either $\alpha$ or $-\alpha$ is positive; }
\item[(3)] {\color{red}the sum and the product of two positive numbers are positive. }
\end{enumerate}

\item[14.] 
From these conditions one derives all the usual rules for manipulation of inequalities. 

\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[15.] 
In particular one finds that every square $\alpha^2$ is either positive or zero; therefore $1 = 1^2$ is a positive number. 

\item[16.] 
By virtue of the order relation the sums $1, 1+1, 1+1+1, \cdots$ are all different. 

\item[17.] 
Hence $\mathbb{R}$ contains the natural numbers, and since it is a field it must contain the subfield formed by all rational numbers. 

\item[18.] 
Finally, $\mathbb{R}$ satisfies the following {\color{blue}completeness} condition: every increasing and bounded sequence of real numbers has a limit. 

\item[19.] 
Let $\alpha_1 < \alpha_2 < \alpha_3 < \cdots < \alpha_n < \cdots$, and assume the existence of a real number $B$ such that $\alpha_n < B$ for all $n$. 

\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[20.] 
Then the completeness condition requires the existence of a number $A = \lim\limits_{n\to\infty} \alpha_n$ with the following property: given any $\varepsilon > 0$ there exists a natural number $n_0$ such that $A - \varepsilon < \alpha_n < A$  for all $n > n_0$.

\item[21.] 
Our discussion of the real-number system is incomplete inasmuch as we have not proved the existence and uniqueness (up to isomorphisms) of a system $\mathbb{R}$ with the postulated properties. 

\item[22.] 
The student who is not thoroughly familiar with {\color{blue}one of the constructive processes by which real numbers can be introduced} should not fail to fill this gap by consulting any textbook in which a full axiomatic treatment of real numbers is given.


\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[23.] 
The equation $x^2 + 1 = 0$ has no solution in $\mathbb{R}$, for $\alpha^2 + 1$ is always positive.

\item[24.] 
{\color{blue}Suppose now that a field $F$ can be found which contains $\mathbb{R}$ as a subfield, and in which the equation $x^2 + 1 = 0$ can be solved. }

\item[25.] 
Denote a solution by $i$. 

\item[26.] 
Then $x^2 + 1 = (x + i)(x - i)$, and the equation $x^2 + 1 = 0$ has exactly two roots in $F$, $i$ and $-i$. 


\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[27.] 
Let $\mathbb{C}$ be the subset of $F$ consisting of all elements which can be expressed in the form $\alpha + i\beta$ with real $\alpha$ and $\beta$. 

\item[28.] 
This representation is unique, for $\alpha + i\beta = \alpha' + i\beta'$ implies $\alpha - \alpha' = -i(\beta - \beta')$; hence $(\alpha - \alpha')^2 = -(\beta - \beta')^2$, and this is possible only if $\alpha = \alpha'$, $\beta = \beta'$. 

\item[29.] 
{\color{blue}The subset $\mathbb{C}$ is a subfield of $F$. } 
 
\item[30.] 
In fact, except for trivial verifications which the reader is asked to carry out, this is exactly what was shown in Sec. 1.1. 


\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[31.] 
What is more, the structure of $\mathbb{C}$ is independent of $F$. 

\item[32.] 
For if $F'$ is another field containing $\mathbb{R}$ and a root $i'$ of the equation $x^2 + 1 = 0$, the corresponding subset $\mathbb{C}'$ is formed by all elements $\alpha + i'\beta$. 

\item[33.] 
There is a one-to-one correspondence between $\mathbb{C}$ and $\mathbb{C}'$ which associates $\alpha + i\beta$ and $\alpha + i'\beta$, and this correspondence is evidently a field isomorphism. 

\item[34.] 
It is thus demonstrated that $\mathbb{C}$ and $\mathbb{C}'$ are isomorphic.


\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[35.] 
We now define the field of complex numbers to be the subfield $\mathbb{C}$ of an arbitrarily given $F$.

\item[36.] 
{\color{blue}We have just seen that the choice of $F$ makes no difference, but we have not yet  shown that there exists a field $F$ with the required properties. }

\item[37.] 
In order to give our definition a meaning it remains to exhibit a field $F$ which contains $\mathbb{R}$ (or a subfield isomorphic with $\mathbb{R}$) and in which the equation $x^2 + 1 = 0$ has a root. 

\item[38.] 
There are many ways in which such a field can be constructed. 


\end{enumerate}

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\begin{frame}{1.3. Justification. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[39.] 
The simplest and most direct method is the following: Consider all expressions of the form $\alpha + i\beta$ where $\alpha, \beta$ are real numbers while the signs $+$ and $i$ are pure symbols ( $+$ does not indicate addition, and $i$ is not an element of a field). 

\item[40.] 
These expressions are elements of a field $F$ in which addition and multiplication are defined by (1) and (2) (observe the two different meanings of the sign $+$). 

\item[41.] 
The elements of the particular form $\alpha + i0$ are seen to constitute a subfield isomorphic to $\mathbb{R}$, and the element $0 + i1$ satisfies the equation $x^2 + 1 = 0$; we obtain in fact $(0 + i1)^2 = -(1 + i0)$. 


\end{enumerate}

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\begin{enumerate}

\item[42.] 
The field $F$ has thus the required properties; moreover, it is identical with the corresponding subfield $\mathbb{C}$, for we can write 
\begin{equation*}
\alpha + i\beta = (\alpha + i0) + \beta(0 + i1).
%\label{eq-}
\end{equation*}

\item[43.] 
{\color{red}The existence of the complex-number field is now proved,} and we can go back to the simpler notation $\alpha + i\beta$ where the $+$ indicates addition in $\mathbb{C}$ and $i$ is a root of the equation $x^2 + 1 = 0$.


\end{enumerate}

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\begin{frame}{1.3. Justification. Exercise - 1 \hfill 作业1A-3 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the system of all matrices of the special form 
$$
\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}
$$
combined by matrix addition and matrix multiplication, is isomorphic to
the field of complex numbers.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.3. Justification. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the complex-number system can be thought of as the field of all polynomials with real coefficients modulo the irreducible polynomial $x^2+1$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.4. Conjugation, Absolute Value. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.]  
A complex number can be denoted either by a single letter $a$, representing an element of the field $\mathbb{C}$, or in the form $\alpha + i\beta$ with real $\alpha$ and $\beta$. 

\item[2.]  
Other standard notations are $z = x + iy$, $\zeta = \xi + i\eta$, $w = u + iv$, and when used in this connection it is tacitly understood that $x, y, \xi, \eta, u, v$ are real numbers. 

\item[3.]  
The real and imaginary part of a complex number $a$ will also be denoted by $\mathrm{Re}\, a$, $\mathrm{Im}\, a$. 

\item[4.]  
In deriving the rules for complex addition and multiplication we used only the fact that $i^2 = -1$. 

\end{enumerate}

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\begin{frame}{1.4. Conjugation, Absolute Value. }

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\begin{enumerate}


\item[5.]  
Since $-i$ has the same property, all rules must remain valid if $i$ is everywhere replaced by $-i$. 

\item[6.]  
Direct verification shows that this is indeed so.

\item[7.]  
The transformation which replaces $\alpha + i\beta$ by $\alpha -i\beta$ is called {\color{blue}complex conjugation}, and $\alpha - i\beta$ is the conjugate of $\alpha+ i\beta$. 

\item[8.]  
The conjugate of $a$ is denoted by $\bar{a}$.

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.]  
A number is real if and only if it is equal to its conjugate. 

\item[10.]  
The conjugation is an involutory transformation: this means that $\bar{\bar{a}} = a$. 

\item[11.]  
{\color{red}The formulas 
\begin{equation*}
\mathrm{Re}\, a = \frac{a+\bar{a}}{2}, \,\,
\mathrm{Im}\, a = \frac{a-\bar{a}}{2i}
\end{equation*}
express the real and imaginary part in terms of the complex number and its conjugate. }

\item[12.]  
By systematic use of the notations $a$ and $\bar{a}$ it is hence possible to {\color{blue}dispense with} the use of separate letters for the real and imaginary part. 

\item[13.]  
It is more convenient, though, to make free use of both notations. 

\end{enumerate}

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\begin{frame}{1.4. Conjugation, Absolute Value. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[14.]  
The fundamental property of conjugation is the one already referred to, namely, that
\begin{equation*}
\begin{aligned}
\overline{a+b} &= \bar{a} + \bar{b} \\ 
\overline{ab} &= \bar{a}\cdot \bar{b}.
\end{aligned}
\end{equation*}

\item[15.]  
The corresponding property for quotients is a consequence: if $ax = b$, then $\bar{a}\bar{x} = \bar{b}$, and hence $\overline{b/a} = \bar{b}/\bar{a}$. 

\item[16.]  
More generally, let $R(a,b,c,\cdots )$ stand for any rational operation applied to the complex numbers $a, b, c, \cdots$. 

\item[17.]  
Then 
\begin{equation*}
\overline{R(a,b,c,\cdots)} = R(\bar{a},\bar{b},\bar{c}, \cdots).
\end{equation*}

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[18.]  
As an application, consider the equation
\begin{equation*}
c_0z^n + c_1z^{n-1} + \cdots + c_{n-1}z + c_n = 0.
\end{equation*}

\item[19.]  
If $\zeta$ is a root of this equation, then $\bar{\zeta}$ is a root of the equation
\begin{equation*}
\bar{c}_0z^n + \bar{c}_1z^{n-1} + \cdots + \bar{c}_{n-1}z + \bar{c}_n = 0.
\end{equation*}

\item[20.]  
In particular, if the coefficients are real, $\zeta$ and $\bar{\zeta}$ are roots of the same equation, and we have the familiar theorem that the nonreal roots of an equation with real coefficients occur in pairs of conjugate roots.

\item[21.]  
The product $a\bar{a} = \alpha^2 + \beta^2$ is always positive or zero. 

\item[22.]  
Its nonnegative square root is called {\color{blue}the modulus or absolute value} of the complex number $a$; it is denoted by $|a|$.

\end{enumerate}

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\begin{enumerate}

\item[23.]  
The terminology and notation are justified by the fact that the modulus of a real number coincides with its numerical value taken with the positive sign.

\item[24.]  
We repeat the definition
\begin{equation*}
a\bar{a} = |a|^2,
\end{equation*}
where $|a|\ge 0$, and observe that $|\bar{a}| = |a|$.

\item[25.]  
For the absolute value of a product we obtain
\begin{equation*}
|ab|^2 = ab\cdot \overline{ab} = ab\bar{a}\bar{b} = a\bar{a}b\bar{b} = |a|^2|b|^2, 
\end{equation*}
and hence
\begin{equation*}
|ab| = |a|\cdot |b|
\end{equation*}
since both are $\ge 0$.

\end{enumerate}

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\begin{frame}{1.4. Conjugation, Absolute Value. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[26.]  
In words: The absolute value of a product is equal to the product of the absolute values of the factors. 

\item[27.]  
It is clear that this property extends to arbitrary finite products:
\begin{equation*}
|a_1a_2 \cdot a_n| = |a_1| \cdot |a_2| \cdots |a_n|. 
\end{equation*}

\item[28.]  
The quotient $a/b, b\neq 0$, satisfies $b(a/b) = a$, and hence we have also 
$|b|\cdot |a/b| = |a|$, or
\begin{equation*}
\left\vert\frac{a}{b}\right\vert = \frac{|a|}{|b|}. 
\end{equation*}

\end{enumerate}

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\begin{frame}{1.4. Conjugation, Absolute Value. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[29.]  
The formula for the absolute value of a sum is not as simple. 

\item[30.]  
We find
\begin{equation*}
|a+b|^2 = (a+b)(\bar{a}+\bar{b}) = a\bar{a}+(a\bar{b}+b\bar{a}) + b\bar{b}
\end{equation*}
or
\begin{equation}
|a+b|^2 = |a|^2 + |b|^2 + 2\mathrm{Re}\,(a\bar{b}). 
\label{eq-7}
\end{equation}

\item[31.]  
The corresponding formula for the difference is 
\begin{equation*}
|a-b|^2 = |a|^2 + |b|^2 - 2\mathrm{Re}\,(a\bar{b}), 
%\label{eq-}
\end{equation*}
and by addition we obtain the identity
\begin{equation}
|a+b|^2 + |a-b|^2= 2(|a|^2 + |b|^2). 
\label{eq-8}
\end{equation}

\end{enumerate}

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\begin{frame}{1.4. Conjugation, Absolute Value. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Verify by calculation that the values of
$$
\frac{z}{z^2+1}
$$
for $z = x + iy$ and $z = x - iy$ are conjugate. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.4. Conjugation, Absolute Value. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the absolute values of
$$
-2i(3+i)(2+4i)(1+i)
\,\,\, \mathrm{and}\,\,\,
\frac{(3+4i)(-1+2i)}{(-1-i)(3-i)}
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.4. Conjugation, Absolute Value. Exercise - 3 \hfill 作业1B-1 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that
$$
\left\vert \frac{a-b}{1-\bar{a}b}\right\vert = 1
$$
if either $|a| = 1$ or $|b| = 1$. 
What exception must be made if $|a| = |b| = 1$?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.4. Conjugation, Absolute Value. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the conditions under which the equation $az + b\bar{z} + c = 0$ in one complex  unknown has exactly one solution, and compute that solution.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.4. Conjugation, Absolute Value. Exercise - 5 \hfill 作业1A-4 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove Lagrange's identity in the complex form
$$
\left\vert \sum\limits_{i=1}^n a_ib_i \right\vert^2 
= \sum\limits_{i=1}^n |a_i|^2 \sum\limits_{i=1}^n |b_i|^2
- \sum\limits_{1\le i<j\le n} |a_i\bar{b}_j-a_j\bar{b}_i|^2. 
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] 
We shall now prove some important inequalities which will be of constant use. 

\item[2.] 
{\color{blue} It is perhaps well to point out that there is no order relation in the complex-number system, and hence all inequalities must be between real numbers.}

\item[3.] 
From the definition of the absolute value we deduce the inequalities
\begin{equation}
\begin{aligned}
& -|a| \le \mathrm{Re}\,(a) \le |a|, \\ 
& -|a| \le \mathrm{Im}\,(a) \le |a|. 
\end{aligned}
\label{eq-9}
\end{equation}

\item[4.] 
The equality $\mathrm{Re}\,(a) = |a|$ holds if and only if $a$ is real and $\ge 0$.

\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
If (\ref{eq-9}) is applied to (\ref{eq-7}), we obtain 
\begin{equation*}
|a+b|^2 \le (|a| + |b|)^2
%\label{eq-}
\end{equation*}
and hence
\begin{equation}
|a+b| \le |a| + |b|. 
\label{eq-10}
\end{equation}

\item[6.] 
This is called the {\color{blue}triangle inequality} for reasons which will emerge later. 

\item[7.] 
By induction it can be extended to arbitrary sums:
\begin{equation}
|a_1+a_2+\cdots+a_n| \le |a_1| + |a_2| + \cdots + |a_n|. 
\label{eq-11}
\end{equation}

\item[8.] 
The absolute value of a sum is at most equal to the sum of the absolute values of the terms.


\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.] 
The reader is well aware of the importance of the estimate (\ref{eq-11}) in the real case, and we shall find it no less important in the theory of complex numbers. 

\item[10.] 
Let us determine all cases of equality in (\ref{eq-11}).

\item[11.] 
In (\ref{eq-10}) the equality holds if and only if $a\bar{b} \ge 0$ (it is convenient to let $c > 0$ indicate that $c$ is real and positive).

\item[12.] 
If $b\neq 0$ this condition can be written in the form $|b|^2(a/b)\ge 0$, and it is hence equivalent to $a/b \ge 0$. 

\item[13.] 
In the general case we proceed as follows. 

\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[14.] 
Suppose that equality holds in (\ref{eq-11}); then
\begin{equation*}
\begin{aligned}
|a_1| + |a_2| + |a_3| + \cdots + |a_n| 
& = |(a_1 + a_2) + a_3 + \cdots + a_n| \\
& \le |a_1 + a_2| + |a_3| + \cdots + |a_n| \\ 
& \le |a_1| + |a_2| + |a_3| + \cdots + |a_n|.
\end{aligned}
%\label{eq-}
\end{equation*}

\item[15.] 
Hence $|a_1 + a_2| = |a_1| + |a_2|$, and if $a_2\neq 0$ we conclude that $a_1/a_2\ge 0$. 

\item[16.] 
But the numbering of the terms is arbitrary; {\color{red}thus the ratio of any two nonzero terms must be positive}. 

\item[17.] 
Suppose conversely that this condition is fulfilled.

\end{enumerate}

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\begin{enumerate}

\item[18.] 
Assuming that $a_1\neq 0$ we obtain
\begin{equation*}
\begin{aligned}
|a_1 + a_2 + \cdots + a_n| 
& = |a_1|\cdot \left\vert 1 + \frac{a_2}{a_1} + \cdots + \frac{a_n}{a_1}\right\vert \\
& = |a_1|\cdot \left( 1 + \frac{a_2}{a_1} + \cdots + \frac{a_n}{a_1}\right) \\
& = |a_1|\cdot \left( 1 + \frac{|a_2|}{|a_1|} + \cdots + \frac{|a_n|}{|a_1|}\right) \\
& = |a_1| + |a_2| + \cdots + |a_n|.
\end{aligned}
%\label{eq-}
\end{equation*}

\end{enumerate}

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\begin{frame}{1.5. Inequalities. \hfill 作业1A-5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[19.] 
{\color{red} To sum up: the sign of equality holds in %(\ref{eq-11}) 
\begin{equation*}
|a_1+a_2+\cdots+a_n| \le |a_1| + |a_2| + \cdots + |a_n|. 
%\label{eq-11}
\end{equation*}
if and only if the ratio of any two nonzero terms is positive. 
}

\item[20.] 
By (\ref{eq-10}) we have also 
\begin{equation*}
|a| = |(a-b)+b| \le |a-b| + |b|
%\label{eq-}
\end{equation*}
or
\begin{equation*}
|a-b| \le |a| - |b|.
%\label{eq-}
\end{equation*}

\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[21.] 
For the same reason $|b|-|a|\le |a-b|$, and these inequalities can be combined to
\begin{equation}
|a-b| \ge \left\vert |a| - |b| \right\vert.
\label{eq-12}
\end{equation}

\item[22.] 
Of course the same estimate can be applied to $|a + b|$.

\item[23.] 
A special case of (\ref{eq-10}) is the inequality
\begin{equation}
|\alpha + i\beta| \le |\alpha| + |\beta|.
\label{eq-13}
\end{equation}
which expresses that the absolute value of a complex number is at most equal to the sum of the absolute values of the real and imaginary part.


\end{enumerate}

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\begin{frame}{1.5. Inequalities. \hfill 作业1A-6 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[24.] 
Many other inequalities whose proof is less immediate are also of frequent use.

\item[25.] 
{\color{red}
Foremost is {\color{blue}Cauchy's inequality} which states that
\begin{equation*}
|a_1b_1+\cdots+a_nb_n|^2 \le (|a_1|^2+\cdots+|a_n|^2)(|b_1|^2+\cdots+|b_n|^2),
%\label{eq-7}
\end{equation*}
or, in shorter notation, 
\begin{equation}
\left\vert \sum\limits_{i=1}^{n} a_ib_i \right\vert^2 
\le \sum\limits_{i=1}^{n} |a_i|^2 \sum\limits_{i=1}^{n} |b_i|^2. 
\label{eq-14}
\end{equation}
}

\item[26.] 
To prove it, let $\lambda$ denote an arbitrary complex number. 

\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[27.] 
We obtain by (\ref{eq-7}) 
\begin{equation}
\sum\limits_{i=1}^{n} |a_i-\lambda \bar{b}_i|^2 = 
\sum\limits_{i=1}^{n} |a_i|^2 + |\lambda|^2 \sum\limits_{i=1}^{n} |b_i|^2 
- 2\mathrm{Re} \left( \bar{\lambda} \sum\limits_{i=1}^{n} a_ib_i \right). 
\label{eq-15}
\end{equation}

\item[28.] 
This expression is $\ge 0$ for all $\lambda$.

\item[29.] 
We can choose
\begin{equation*}
\lambda = \frac{\sum\limits_{i=1}^{n} a_ib_i}{\sum\limits_{i=1}^{n} |b_i|^2},
%\label{eq-}
\end{equation*}
for if the denominator should vanish there is nothing to prove. 

\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[30.] 
{\color{blue}This choice is not arbitrary, but it is dictated by the desire to make the expression (\ref{eq-15}) as small as possible.}

\item[31.] 
Substituting in (\ref{eq-15}) we find, after simplifications, 
\begin{equation*}
\sum\limits_{i=1}^{n} |a_i|^2 
- \frac{\left\vert \sum\limits_{i=1}^{n} a_ib_i \right\vert ^2 }
{\sum\limits_{i=1}^{n} |b_i|^2 }
\ge 0
%\label{eq-}
\end{equation*}
which is equivalent to (\ref{eq-14}).


\end{enumerate}

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\begin{frame}{1.5. Inequalities. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[32.] 
From (\ref{eq-15}) we conclude further that the sign of equality holds in (\ref{eq-14}) if and only if the $a_i$ are proportional to the $\bar{b}_i$.

\item[33.] 
Cauchy's inequality can also be proved by means of Lagrange's identity (Sec. 1.4, Ex. 4).


\end{enumerate}

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\begin{frame}{1.5. Inequalities. Exercise - 1 \hfill 作业1B-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that
$$
\left\vert \frac{a-b}{1-\bar{a}b} \right\vert < 1
$$
if $|a| < 1$ and $|b| < 1$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.5. Inequalities. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove Cauchy's inequality by induction.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.5. Inequalities. Exercise - 3}

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\begin{itemize}
\item  {\color{red}Question. 
If $|a_i| < 1$, $\lambda_i\ge 0$ for $i = 1,\cdots,n$, and 
$\lambda_1 + \lambda_2 + \cdots + \lambda_n = 1$,
show that
$$
| \lambda_1a_1 + \lambda_2a_2 + \cdots + \lambda_na_n | < 1. 
$$
}

\item  Answer. 
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\begin{frame}{1.5. Inequalities. Exercise - 4}

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\begin{itemize}
\item  {\color{red}Question. 
Show that there are complex numbers $z$ satisfying
$$
|z-a|+|z+a|=2|c| 
$$
if and only if $|a| \le |c|$. If this condition is fulfilled, what are the smallest
and largest values of $|z|$?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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